in reply to Escaping a variable

my $foo = "4D";
my $bar = "\x4D";

print $bar . "\n";

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Doesn't produce

Illegal hexadecimal digit '$' ignored at roman.pl line 10.
4D.

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Why don't you please edit your post and include the actual code that is giving you that error message? I suspect I know what it looks like, but it's your job to actually show code that exhibits the problem you're describing.

At any rate, hex will, given the string contained in $foo as a parameter, return the base ten integer value 77.


Dave

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Re^2: Escaping a variable
by Draxter (Initiate) on Nov 08, 2011 at 09:19 UTC

    Hi,

    Sorry, you are right of course.

    This is what I do: 

    my $foo = "4D";
my $bar = "\x$foo";

print $bar . "\n";

Illegal hexadecimal digit '$' ignored at roman.pl line 7.
4D

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[reply]
[d/l]

      print hex($foo), "\n";

      The output:

      77

      [download]

      Dave

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[d/l]
[select]

        I'll be more specific. How can I get $string and $bar to be the same by the end of this code: 

        #!/usr/bin/perl

use warnings;
use strict;

my $bar = "\x04\x4d";
my $foo = "044d";
my $string;

while ($foo =~ m/(..)/g) {
        $string .= chr($1);
}

print "string: $string\n";
print "bar: $bar\n";

Argument "4d" isn't numeric in chr at roman.pl line 11.
string: 
bar: M

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        Thanks,

        Eran.

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[d/l]
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[d/l]

        I don't want the hex value of the variable, I need the variable to be printed as an ascii representation of the hex value.

[reply]

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